@vbhvgupta Please give the options atleast.... :D
@vbhvgupta said:/u/LogrhythmTHe sum of first 4 terms of ap is 28 and sum of first 8 term of the same ap is 88. Find the sum of first 16 terms of the ap?
4/2[2a + 3d] = 28
8/2[2a + 7d] = 88
solving simultaneously, d = 2; hence a =4
sum of first 16 terms - 16/2[8 + 15*2] = 304
@vbhvgupta said:@Logrhythm can you solve this?
304?
4/2(2a+3d)=28....i
8/2(2a+7d)=88....ii
solve i and ii
a=4,d=2
16/2(2*4+15*2)=304
@Subhashdec2 said:120+2*120(4/5 + 4^2/5^2+.....)120+2*120(4)120*9=1080
could you explain ? all I know is sum of infinite gp is a[1/(1-r)]
@VaibhavJ when the ball drops height is 120m.
total height= 120 + 120*4/5 (1st rebound) + 120*4/5(balls falls back after first rebound due to gravitational force...loll..) + 120*4/5*4/5(2nd rebound) + 120*4/5*4/5( gravitational force comes into play again) + ..... infinity
=> 120+ 120(4/5 + 4/5 + 4/5*4/5 + 4/5*4/5 +..... )
=> 120 + 120*2(4/5+4/5*4/5 + (4/5)^3 + ...)
=> 120 + 240 (4/5*5)
=> 120 + 960
=> 1080 answer/....
hope it helps..!!! :)
@jai.tuteja said:@VaibhavJ when the ball drops height is 120m.total height= 120 + 120*4/5 (1st rebound) + 120*4/5(balls falls back after first rebound due to gravitational force...loll..) + 120*4/5*4/5(2nd rebound) + 120*4/5*4/5( gravitational force comes into play again) + ..... infinity=> 120+ 120(4/5 + 4/5 + 4/5*4/5 + 4/5*4/5 +..... )=> 120 + 120*2(4/5+4/5*4/5 + (4/5)^3 + ...)=> 120 + 240 (4/5*5)=> 120 + 960=> 1080 answer/....hope it helps..!!!
Understood. Thanks a lot :)
@vbhvgupta said:Q
120 + 2* 120*4/5 + 2*120*4/5 *4/5 +..... = 120 + 2* 120 [ 4/5 + (4/5)^2+...] = 120+ 240*4 = 1080
After the first bounce total distance for second bounce is going up+coming down [ think of a practical case]...
and 4/5 + (4/5)^2+... is a infinite g.p series case
After the first bounce total distance for second bounce is going up+coming down [ think of a practical case]...
and 4/5 + (4/5)^2+... is a infinite g.p series case
@vbhvgupta said:if a times the ath term of an ap is equal to the b times the bth term, find (a+b)th term?1. 0 3. a^2 + b^22. a^2 - b^2 4. a - b5. 1
make equations and solve for d.
d=-a
therefore, (a+b)th term =0
@sunnychopra89 said:What is the no. of +ve integer triplets(a,b,c) that satisfy abc=4(a+b+c) with a
zero??
Let N = 111...111(73 times). When N is divided by 259, the remainder is R1, and when N is divided by 32, the remainder is R2. Then R1 + R2 is equal to
A. 6
B. 8
C. 253
D. None of these..
Kindly share the approach ...
Thank you.. :)
A. 6
B. 8
C. 253
D. None of these..
Kindly share the approach ...
Thank you.. :)
@TootaHuaDil said:Let N = 111...111(73 times). When N is divided by 259, the remainder is R1, and when N is divided by 32, the remainder is R2. Then R1 + R2 is equal toA. 6 B. 8 C. 253 D. None of these..Kindly share the approach ...Thank you..
none of these ?
@TootaHuaDil said:Let N = 111...111(73 times). When N is divided by 259, the remainder is R1, and when N is divided by 32, the remainder is R2. Then R1 + R2 is equal toA. 6 B. 8 C. 253 D. None of these..Kindly share the approach ...Thank you..
259=7*37
1111....73 times mod 7 =1
1111...73 mod 37=1
7a+1=37b+1
a=37
remainder =37*7+1 mod 259=1
r1=1
for 32
11111 mod 32=7
8??
@VaibhavJ said:1] 1-2-3 = -4 2] 2-3-4 = -5we see a patternsummation of this will be -[33/2(4 + {33-1}1] = -650also we see that 100th term will be +34therefore summation - -650+34 = -626is this correct ?
yes, i is -626
@Subhashdec2 said:259=7*371111....73 times mod 7 =11111...73 mod 37=17a+1=37b+1a=37remainder =37*7+1 mod 259=1r1=1for 3211111 mod 32=78??
yes, it'll be 8
Q2